Question

help me prove this identity plz

Answer 1

\(\dfrac{\sin 2x}{\sin x} - \dfrac{\cos 2x}{\cos x} = \sec x\)

Double angle identities:

\(\sin 2x = 2 \sin x \cos x\)

\(\cos 2x = 2 \cos^2 x - 1\)

\(\dfrac{2 \sin x \cos x}{\sin x} - \dfrac{2 \cos^2 x - 1}{\cos x} = \sec x\)

\(\dfrac{2 \sin x \cos x}{\sin x} - \dfrac{2 \cos^2 x}{\cos x} + \dfrac{1}{\cos x} = \sec x\)

\(\dfrac{2 \cancel{\sin x} \cos x}{\cancel{\sin x}~1} - \dfrac{2 \cos^\cancel{2} x}{\cancel{\cos x}~1} + \dfrac{1}{\cos x} = \sec x\)

Answer 2

Now just simplify the left side and use the definition of secant.